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Solve 2^x+3^x-4^x+6^x-9^x=1 Over the Reals | IOQM

JEE Maths question with a full step-by-step solution.

Question
The number of all real numbers xx satisfying 2x+3x−4x+6x−9x=12^x + 3^x - 4^x + 6^x - 9^x = 1.
Solution
Answer: 0
Step 1: Let p=2xp = 2^x and q=3xq = 3^x, both positive. Then 4x=p24^x = p^2, 9x=q29^x = q^2, 6x=pq6^x = pq, and the equation becomes
p+q−p2+pq−q2=1.p + q - p^2 + pq - q^2 = 1.
 Step 2: Rearrange into a sum of squares. Bring all terms to one side,
p2+q2−pq−p−q+1=0,p^2 + q^2 - pq - p - q + 1 = 0,
 and multiply by 22:
(p−q)2+(q−1)2+(1−p)2=0.(p - q)^2 + (q - 1)^2 + (1 - p)^2 = 0.
 Step 3: Each square must vanish: p=qp = q, q=1q = 1, p=1p = 1, so p=q=1p = q = 1. Step 4: Then 2x=12^x = 1 and 3x=13^x = 1, giving x=0x = 0. Answer: x=0x = 0 is the only real solution.
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